I'm trying to calculate the Energy you would get in a fusion reactor from the fusion of deuterium and tritium:
${}^2H+{}^3H \rightarrow {}^4He + n$
Using this Equation:
$E = E_{rest} + E_{kin} = mc^2 + \frac12mv^2$
And these values i found online:
$m_{Deuterium} \approx 2.01410177811u$
$m_{Tritium} \approx 3.01604928u$
$m_{Helium4} \approx 4.002603254u$
$m_{Neutron} \approx 1.03352196257794u$
These velocities are at ~100 million Kelvin
$v_{Deuterium} \approx 1500\frac{km}s$
$v_{Tritium} \approx 1000\frac{km}s$
Plugging in the values i get this:
$E_{Deuterium} \approx 1882.3819988MeV$
$E_{Tritium} \approx 2819.97477352MeV$
$E_{Helium4} \approx 3728.40131MeV$
$E_{Neutron} \approx 962.719610361MeV$
Then the Energy before the reaction minus the energy after the reaction is:
$\Delta E \approx 10.84669MeV$
But on the Wikipedia about fusion it says that the reaction should release $17.59MeV$ in kinetic energy.
I assume the problem could be the inaccurate velocities, but I'm not sure the difference would be so big.
Best Answer
Kinetic energies do not have to be taken into account: they only serve to overcome Coulomb repulsion. In the Sun, fusions occur at "low temperature". Do the usual Q checkup with the masses and you'll easily find the correct answer.